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Solving matrix equation Ax=B

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data

    You are given matrices A and B:

    1 -2 1
    A = 0 0 0
    0 1 1


    1
    B = 0
    2

    Solve the equation Ax=B

    2. Relevant equations

    x=A^(-1)*B

    3. The attempt at a solution

    I carried out exactly the steps required for inverting a matrix. When got to th part where the determinant is calculated, I obtained a zero determinant. I thought this meant that the inverse matrix could not be calculated. This is also the result that an online matrix calculator gave for A. Hence I anwered "unsolvable" in the exam. Yet I was only given 1 out of 3 points.

    What should I have done differently?

    Thanks for your help!
     
  2. jcsd
  3. Dec 16, 2013 #2

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    The fact that A has no inverse means that there exist some vectors B such that
    Ax = B
    is not solvable. However as long as B is in the image of A it is solvable. In particular for this problem notice that A always forces the second coordinate of Ax to be equal to zero - so if B had a nonzero second coordinate the problem would be unsolvable. Luckily B has zero for its second coordinate, so it will turn out that Ax=B is solvable in this case.

    As far as figuring out what x can be equal to, you can write down a system of equations from Ax=B which you should be able to solve. Try that and show us what you get!
     
  4. Dec 16, 2013 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    For any nxn linear system, exactly one of the following statements is true: Ax = b has (i) exactly one solution; (ii) no solutions; or (iii) infinitely many solutions. You chose (ii), but without checking the details. The easiest way is to just write out the equations:
    [tex]1x -2y + 1x = 1\\
    0x+0y+0z=0\\
    0x+1y+1z = 2[/tex]
    The second equation is redundant, so we just have two equations in three unknowns:
    [tex] \begin{array}{rc} x - 2y + z & = 1\\
    y + z &= 2 \end{array}
    [/tex]
     
  5. Dec 17, 2013 #4
    Thanks alot guys! That's actually all familiar stuff to me and it all makes sense now, but for some reason I just can't think straight when it comes to matrices. Thanks again!
     
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